Question

If $k$ be the perimeter of the $\Delta ABC$ then $b{\cos }^2\frac{C}{2}+c{\cos }^2\frac{B}{2}$ is equal to

• A. $k$
• B. $2k$
• C. $\frac{k}{2}$
• D. none of these

Answer 1

Answer: (C)

$\frac{k}{2}$

We know, $1+\cos \left(\theta \right)=2{\cos }^2\left(\frac{\theta }{2}\right)$

Hence, $b\left({\cos }^2\left(\frac{C}{2}\right)\right)+c\left({\cos }^2\left(\frac{B}{2}\right)\right)$

$=\frac{b(1+\cos C)+c(1+\cos B)}{2}$
$=\frac{1}{2}[b+c+b\cos C+c\cos B]$
$=\frac{1}{2}[b+c+a]$
$=\frac{k}{2}$